用C语言编写在一个字符串中找出元音字母a,e,i,o,u出现的次数。 需要区分 大小写！！ 只统计小写元音字 c语言，统计字符串中个元音字母（即a、e、i、o、u）的个数...
C\u8bed\u8a00\uff1a\u7f16\u5199\u7a0b\u5e8f\u7edf\u8ba1\u4ece\u952e\u76d8\u8f93\u5165\u7684\u5b57\u7b26\u4e32s\u4e2d\u5404\u5143\u97f3\u5b57\u6bcd\uff08\u5373A.E.U.I.O\uff09\u7684\u4e2a\u6570\uff0c\u5176\u4e2d\u5b57\u6bcd\u4e0d\u5206\u5927#include "stdio.h"
int count(char c[], char p)
{
int i, count = 0;
for (i = 0; c[i] != '\0'; i++)
if (c[i] == p || c[i] == p + 32)
count++;
return count;
}
int main(void)
{
int i;
char c[50];
int b[5] = { 0 };
gets(c);
printf("A\u4e2a\u6570%d\n", count(c, 'A'));
printf("E\u4e2a\u6570%d\n", count(c, 'E'));
printf("I\u4e2a\u6570%d\n", count(c, 'I'));
printf("O\u4e2a\u6570%d\n", count(c, 'O'));
printf("U\u4e2a\u6570%d\n", count(c, 'U' ));
return 0;
}
void fun (char *s , int num[5] )
{
int k , i = 5 ; //i = 5 \u662f\u6307\u5143\u97f3\u7684\u4e2a\u6570\uff0c\u4e5f\u5c31\u662fnum[0]\u8868\u793aa\u7684\u4e2a\u6570\uff0c\u4ee5\u6b64\u7c7b\u63a8
for (k=0 ; k<i ; k++) //k\u53ea\u662f\u4e2a\u8ba1\u6570
num[k ] = 0 ;
for ( ; *s ; s++) //s\u9996\u5148\u6307\u5411\u5b57\u7b26\u4e32\u7684\u7b2c\u4e00\u4e2a\u5b57\u7b26
{
i= -1 ; //
switch(*s) //\u5224\u65ads\u6307\u5411\u7684\u5b57\u7b26\u662f\u5426\u4e3a\u5143\u97f3
{ ... //\u8fd9\u91cc\uff0c\u5982\u679c*s==a , i = 0\uff0c\u4ee5\u6b64\u7c7b\u63a8\uff0c\u5982\u679c*s\u4e0d\u662f\u5143\u97f3\uff0ci = -1;
}
if ( i>=0)
num[i] ++;
}
}
#include <stdio.h>
void HowVowel(int *pr,char *ps)
{
char v[7]="aeiou";
for(;*ps;ps++)
{
switch(*ps)
{
case 'a':
++pr[0];
break;
case 'e':
++pr[1];
break;
case 'i':
++pr[2];
break;
case 'o':
++pr[3];
break;
case 'u':
++pr[4];
break;
default:
break;
}
}
}
int main(void)
{
char Str[200]="";
int Result[5]={0,},i;
gets(Str);
HowVowel(Result,Str);
for(i=0;i<5;printf("%d ",Result[i++]));
printf("
");
return 0;
}
在你的代码上的修改，去掉|0x20，这个是包含大小写的；
void HowVowel(int *pr,char *ps)
{
char v[7]="aeiou",i;
for(;*ps;ps++)
for(i=0;*(v+i);i++)
if((*ps)==*(v+i))
pr[i]++;
}
修改一下：
void HowVowel(int *pr,char *ps)
{
char v[]="aeiou",i;
int count[26]={0}
for(;*ps;ps++)
if(*ps<='z' && *ps>='a')count[*ps-'a']++;各个字母
for(i=0;v[i];i++)pr[i]=count[v[i]-'a'];
}
只要小写的话
if(*ps==*(v+i))
if((*ps|0x20)==*(v+i))
是一个取巧的办法，可以将‘A'-’Z'转换为'a'-'z'
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